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Vector a=(1,2), b=(1,-1), then /a+2b/=
A+2b =(1,2)+2*(1,-1)=(3,0)
So |a+2b|=√(3 2+0 2)
= 3
Given vector a=(-1,2), b=(1,1), then 3a-2b=? Given the vector a=(-1,2), b=(1,1), then 3a-2b=?
Parse
3A-2b=3(-1 2)-2(1 1)
=(-3 6)-(2 2)
=(-3-2 6-2)
=(-5 4)
Children's shoes to adopt
Given vector a=(3, root number 3), b=(1,0), then (a-2b) b=
Isn't the operation of a vector a vector at the end?
Given a vector =(root number 3,1), b =(0,-1), c =(K, root number 3, if a-2b is collinear with c to find K
A-2b =(root number 3,3)
Because c=(K, root 3) is collinear with a-2b
So 3K = root 3x root 3, K =1
A-2b=(root 3,3)
Because c=(K, root 3) is collinear with a-2b
So 3K = root 3x root 3, K =1
In the triangle ABC, A=Section/6,(1+Root 3) c=2b, find C.(2) If the vector CBXCA=1+Root 3, find a, b, c In the triangle ABC, A=Section/6,(1+Root 3) c=2b. find C.(2) If the vector CBXCA=1+Root 3, find a, b, c In the triangle ABC, A=pa/6,(1+Root 3) c=2b.
A=30°, B+C=150°, then by sine theorem (1+Root 3) C=2b,(1+Root 3) sinC=2sinB, i.e.,(1+Root 3) sinC=2sin (150°-C), C=45°, A=30°, C=45°, B=105°, then by sine theorem a/sinA=b/sinB=c/sinC (sin75=(Root 2+Root 6...
In △ABC, the opposite sides of ABC are abc, A=π/6,(1+Root 3) c=2b.
(1) C=π/4
(2) Is the CB vector (1,√3)?
If a=2, b=√2 6, c=2√2
Process (I only know stupid methods):
(1) From cosine theorem, a^2=b^2+c^2-2bc*cos (π/6)1
C^2= a^2+b^2-2ab*cosC2
B =(1 3)/2*c3
Solve the equation group to obtain C=π/4, c=√2a, b=(√2 6)/2*a
(2) If CB vector is (1,√3) then a^2= CB vector ^2=1^2 3^2=4
So a=2 substitute c=√2a, b=(√2 6)/2*a
A=2, b=√2 6, c=2√2
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